# Solving of System of Two Equation with Two Variables. Addition Method

This method is based on facts about equivalent systems.

Example. Solve system of equations {(2x+3y=7),(3x-y=16):}.

Multiplying both sides of second equation by 3, we obtain equivalent system: {(2x+3y=7),(9x-3y=48):}.

Now, if we replace second equation (we can replace first as well) with sum of first and second equations, we will obtain equivalent system: {(2x+3y=7),((2x+3y)+(9x-3y)=7+48):} or {(2x+3y=7),(11x=55):}.

From second equation x=5. Plugging this value in the first equation, we obtain that 2*5+3y=7 or y=-1.

Thus, (5;-1) is solution of the system.

Now, let's closer see how this method works. You probably asked why first step was multiplying second equation by 3. It was done, because when we added equations, y was eliminated from second equation.

In general, we want to multiply both sides of equations by such number, that coeffcients near one of the variables were equal (but with opposite signs).

We multiplied second equation by 3, and obtained that coeffcient near y is same as near y in first equation, but with opposite sign.

When you choose what variable to eliminate, choose that way that will lead to less calculations.

In this example it is easier to eliminate y variable, but just for example let's see how to eliminate x.

Coefficients near x are 2 and 3. We can multiply second equation by -3/2 and then coeffcient will be -2, but then calculations will involve fractions.

Instead, we will find least common multiplier of 2 and 3: LCM(2,3)=6.

Now we want to make coeffcient near x to be 6 and -6 respectively. For this multiply first equation by 3 and second by -2:

{(6x+9y=21),(-6x+2y=-32):}. If we now add equations we will eliminate x and obtain that 11y=-11 or y=-1. Then we can find from first equation that x=5. As you see we obtained same solution.

In general, there are many ways to make coefficients near variables equal, but with opposite signs.